Is a group isomorphic to the internal product of its Sylow p-subgroups?

Let $$G$$ be a group such that its order is a product of distinct primes $$p_1, \dots, p_n$$ and let $$P_i$$ denote each Sylow $$p_i$$-subgroup. Is $$P_1 \dots P_n$$ (the internal or Frobenius product) equal to $$G$$, that is, $$G = P_1 \dots P_n$$?

• @ChrisGerig oh, okay! Is there a way to move the question or I should delete it here and re-post it there? Nov 4, 2020 at 0:13
• @YCor Indeed, $G = P_1 \dots P_n$ is more precise. What notation would you suggest for the product instead of $P_1 \dots P_n$? I am happy with it, but I struggle to think of anything better. Nov 5, 2020 at 13:13
• @YCor Indeed. Everything should be correct now. Thanks for the feedback! Nov 5, 2020 at 16:12

For $$G$$ finite, one of the many theorems of P. Hall is that your condition holds whenever $$G$$ is solvable (regardless of order). A note of Rowley and Holt discusses the general problem (and includes a reference for said result of Hall), and provides a few non-solvable examples. They also show that such a product does not exist for the finite simple group $$G=U_3(3)$$. So in full generality the answer is "no". However, if $$|G|$$ is square-free, as you assume, then $$G$$ is solvable (see also these notes for more details), so the answer will be "yes" in this case.
• In the original question $|G|$ is supposed to be a product of distinct primes. The example of Rowley and Holt is interesting and I am glad to have seen it, but their group has order $2^5.3^3.7$. Nov 4, 2020 at 0:42
There is a more direct proof than quoting the fairly deep Theorem of P. Hall, but you do need to know a little transfer theory. The argument that follows is well-known and may be found in many group theory texts. We proceed by induction, there being nothing to prove when $$n = 1$$. Suppose then that $$n > 1$$ and that the result is true for smaller values of $$n$$. If $$|G| = p_{1}p_{2} \ldots p_{n}$$ where $$p_{1} < p_{2} < p_{3} < \ldots < p_{n}$$ are primes, and if we let $$P_{i}$$ be a Sylow $$p_{i}$$-subgroup of $$G$$ for each $$i$$, then we note that the order of $$N_{G}(P_{1})/C_{G}(P_{1})$$ divides $$p_{1}-1.$$ But since $$p_{1}$$ is the smallest prime divisor of $$|G|$$, we see that $$N_{G}(P_{1}) = C_{G}(P_{1})$$.
By Burnside's transfer theorem, $$G$$ has a normal $$p_{1}$$-complement, which means that $$G$$ has a normal subgroup $$H_{1}$$ of order $$p_{2}p_{3} \ldots p_{n}.$$ Then $$H_{1}$$ contains all elements of $$G$$ of order coprime to $$p_{1}$$, and we have $$G = H_{1}P_{1} = P_{1}H_{1}$$, since $$H_{1} \lhd G$$.
By induction, we have $$H_{1} = P_{2} P_{3} \ldots P_{n}$$, so that $$G = P_{1}H_{1} = P_{1}P_{2} \ldots P_{n}.$$